Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + x}{x - 1} = \dfrac{90}{x - 1}$
Answer: Multiply both sides by $x - 1$ $ \dfrac{x^2 + x}{x - 1} (x - 1) = \dfrac{90}{x - 1} (x - 1)$ $ x^2 + x = 90$ Subtract $90$ from both sides: $ x^2 + x - (90) = 90 - (90)$ $ x^2 + x - 90 = 0$ Factor the expression: $ (x + 10)(x - 9) = 0$ Therefore $x = -10$ or $x = 9$ The original expression is defined at $x = -10$ and $x = 9$, so there are no extraneous solutions.